The correct answer is E)52
When there is Hardy-Weinberg equilibrium like in this case of a single locus with two alleles denoted A and a with frequencies f(A) = p and f(a) = q, the expected genotype frequencies under random mating are f(AA) = p² for the AA homozygotes, f(aa) = q² for the aa homozygotes, and f(Aa) = 2pq for the heterozygotes. Let's put that A is dominant and a is recessive allele. In Hardy-Weinberg equilibrium we have:
p²+2*p*q+q²= 1 p+q=1 p=1-q
f(a) =0.4=q q²=0.4²=0.16=16%
p= 1-0.4=0.6 p²=0.6²=0.36=36%
Percentage of a homozygous genotype in the population is 16%+36%=52%
A group of similar cells that perform a single function is called a tissue.
The end result is production of four haploid cells (n chromosomes, 23 in humans) from the two<span> haploid cells (with n chromosomes, each consisting of </span>two<span> sister chromatids) produced in </span>meiosis<span> I. The four main </span>steps of Meiosis II<span> are: Prophase </span>II<span>, Metaphase </span>II<span>, Anaphase </span>II<span>, and Telophase </span>II<span>.</span>
C. employment and unemployment
because it is the only one that makes sense
